# Is this computation of the tensor product correct?

I'm reading the proof of the existence of the tensor product. If $M,N$ are two $R$-modules then we can construct the tensor product $T$ as the quotient $C/D$ where $C$ is the free module over $M \times N$ and $D$ is the submodule generated by the set of all elements in $C$ of the form $$(m+m^\prime, n) - (m,n) - (m^\prime, n)$$ $$(m, n+n^\prime) -(m,n) - (m,n^\prime)$$ $$(am, n) - a(m,n)$$ $$(m,an) - a(m,n)$$

I use $(m,n)$ to denote the element $e_{(m,n)} \in F(M\times N)$. Since $F(S) \cong \bigoplus_{s \in S} R$ I picture these elements as $e_{(m,0)} = (0, \dots, 0,1, 0, \dots)$ where the $1$ here is at position $m$ and $e_{(m,n)}$ the sequence with $1$ at position $m \cdot |M| + n$ and so forth.

Is this correct so far?

Now I wanted to see what this looks like. So I computed the tensor product of $M = N = \mathbb Z / 2 \mathbb Z$ over $R=\mathbb Z$. For $C$ I get that $C \cong \mathbb Z^4$. Then I computed all the elements above and noticed that $D \cong \langle \{(1,0,0,0), (0,1,0,0), (0,0,1,0)\} \rangle$. Hence $$M \otimes N = \mathbb Z / 2 \mathbb Z \otimes_{\mathbb Z} \mathbb Z / 2 \mathbb Z = \langle (0,0,0,1) \rangle \cong \mathbb Z$$

Is this correct?

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It is better to study the tensor product abstractly instead of worrying about the specific construction. If you do that then you get the identity $$R / I \otimes_R M \cong M / IM$$ which should help you here (where $I \lhd R$ is an ideal). It's proved here: math.stackexchange.com/questions/150114/… . –  Paul Slevin Jun 7 '12 at 14:06
Well, the result is definitely wrong: $Z/aZ \otimes_Z Z/bZ =Z/(a,b)Z$ –  awllower Jun 7 '12 at 14:07
Thanks @PaulSlevin. Though I think to complement my studies it's good to look at concrete examples, too. Perhaps I should post my sums then it should become apparent where I messed up. –  Rudy the Reindeer Jun 7 '12 at 14:10
@ClarkKent Tensor products are 1 million percent confusing when you first start learning! –  user38268 Jun 7 '12 at 14:25
@BenjaminLim I linked to the proof:) I'll put it in my answer though. –  Paul Slevin Jun 7 '12 at 14:34

I will add to the comment I gave beneath your question. I know you said that you were interested in the specific construction, but perhaps later you will find this answer useful too.

We know that for any ideal $I \lhd R$, and any $R$-module $M$, we have the isomorphism:

$$R/I \otimes_R M \cong M/IM$$

To see this, take the exact sequence $$0 \to I \to R \to R/I \to 0$$ and tensor by $M$ (recalling that tensoring by $M$ preserves right-exact sequences), then use the standard isomorphism theorem for modules.

So in your case it follows that

$$\mathbb{Z}/2 \mathbb Z \otimes_\mathbb{Z} \mathbb{Z}/2 \mathbb Z \cong \frac{\mathbb{Z}/2\mathbb Z }{ \langle 2 \rangle \mathbb{Z} / 2\mathbb{Z}} \cong \mathbb{Z} / 2 \mathbb{Z}$$

Since the bottom part of the quotient is just $0$.

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Your last isomorphism above cannot hold; the guy on the left is a finite group but the one on the right is 100% not a finite group!

What it should be isomorphic to is $\Bbb{Z}/d\Bbb{Z}$ where $d$ is the greatest common divisor of $2$ and $2$, in this case $2$ itself so that $\Bbb{Z}/2\Bbb{Z} \otimes_\Bbb{Z} \Bbb{Z}/2\Bbb{Z} \cong \Bbb{Z}/2\Bbb{Z}$ . To see this, given any elementary tensor $a \otimes b$ in the tensor product, there are only 4 possibilities: $a$ odd $b$ even, $a$ odd $b$ odd, $a$ even $b$ odd, $a$ odd $b$ even. But the cases where you have an even appearing are just zero because

$$\begin{eqnarray*} 0 \otimes 1 &=& (0 + 0) \otimes 1 \\ &=& 0 \otimes 1 + 0 \otimes 1 \\ \implies 0 \otimes 1 &=& 0 \end{eqnarray*}$$

and similarly $1 \otimes 0 = 0$. Hence there are only two distinct elements that appear, namely $1 \otimes 1$ and $0$ so that your tensor product is isomorphic to the cyclic group of order 2.

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